 Hi, I'm Zor. Welcome to Unizor education. I would like to spend some time basically proving some theorems, which I present as problems on Unizor.com. So these problems are related only to the material in solid geometry, which we have already covered, which basically are lines and planes parallel and perpendicular to each other. So, just a couple of very, very simple theorems, which are presented here right now, and I would like to be as rigorous as possible. So it might actually take, I don't know, maybe a little longer, because the theorems are actually relatively trivial, but I would like to be precise as much as I can. Okay, now the theorem number one. If you have two lines, A and B, which are parallel to each other, and you have a plane, so this is a plane, and these are lines. And you know that one of these lines are perpendicular to the plane. Well, it's kind of obvious that another line also should be perpendicular. So this is your A line. This is your B line. So they are parallel to each other, and I know that A is perpendicular to the plane gamma. Let's connect these points. This is the base of the perpendicular A, and this is the point of intersection. Now, by the way, here is one simple kind of deviation towards precision. I said, okay, B is a point of intersection of line B and plane gamma. Does it exist? Maybe they don't intersect. Well, let's just think about it. If B does not intersect gamma, then B is parallel to gamma. Now, A is parallel to B, and we have already proven before in one of the lectures that if you have two parallel lines and one of them is parallel to a plane, then another is also parallel to that same plane. And if A is parallel to plane, there is no intersection, and there is no intersection cannot be perpendicular, obviously, right? So just a little detail, which in many cases is just completely missed. People are just saying, okay, let B be the intersection point of line B and gamma. We have to think about existence of this. All right, fine. Now, to prove that B, line B, is perpendicular to gamma, I need to prove it's sufficient to prove, again, not I need to prove, it's sufficient to prove another precision point. It is sufficient to prove that line B is perpendicular to at least two lines on the plane, which are passing through the same point of intersection. Well, let's just think about this way. A and B are parallel. From the definition of the parallel lines, they are supposed to line the same plane and have no common points, no intersection. So A and B do lie in the same plane. Well, let's just draw that plane somewhere here and basically points A and B belong to lines A and B correspondingly and the segment A and B would be an inter... belongs to intersection of this plane which goes through A and B and the plane gamma. Why? Again, because point A belongs to both line A and point gamma, right? And therefore A supposed to belong to this plane as well because it goes through the A. So that's why A is an intersection. Same thing B. B is an intersection between line B and plane gamma. Now line B belongs to this big plane. Therefore, point B belongs to both this big plane and gamma which means it belongs to intersection. And again, being as precise as possible, I can remember that if two points belong to intersection of two planes, then the line which goes through them also belongs. So the whole segment A and B belongs to the plane gamma. Now obviously within this plane which now contains this line, this line and this segment as well lines A and B are parallel. Now, I also know that this angle is 90 degrees because A is by the condition of the problem is perpendicular to the plane and therefore it's perpendicular to any line which goes through A. So I have this perpendicularity. Now on the plane, if you have two lines parallel to each other and you have a perpendicular to one line, it will be perpendicular to another. That's from the plane geometry, which we all know. Which means that this angle is also 90 degrees. So line B is perpendicular to segment A and B or line which goes through the segment. Okay. Now that's one. Again, you remember it's sufficient to prove the perpendicularity of the line to a plane. It's sufficient to prove its perpendicularity to two lines on the plane gamma that goes through the point B. We found one line. How about another? Well, it's also kind of simple. Let's draw two parallel lines through A and B. And now let's compare. Let's say this is A prime and this is B prime. Let's compare angle. We need these points as well. C and T somewhere. Angle C, A, A prime and angle D, B, B prime. This angle and this angle. Now, by construction, A, A prime is parallel to B, B prime. By the preposition of the theorem, C, A is parallel to D, B. So these two angles are angles with correspondingly parallel sides. And therefore, they are equal to each other. We know that from the plane geometry, if you have two angles, it doesn't really matter whether they are in space, separated or in the same plane. We can always bring them to the same plane if you really want to. But as long as the parallel lines, the sides are parallel, then the angles are equal. So this one is 90 degree because the line perpendicular to any line, A is perpendicular to any line on the plane, which goes through the point A. And therefore, this is also supposed to be 90 degrees. This is my second line. Well, actually, kind of similarly, instead of using this line, A, B, I can immediately draw two lines here and two parallel, correspondingly parallel lines here and just deal with those angles. It doesn't really matter. Okay, so that actually proves that line B is perpendicular to the plane gamma. We found two lines on the gamma, which goes through the point of intersection B, which are perpendicular to the line B. That's why there is a perpendicularity. Okay, that's my first theorem. The second theorem is kind of an opposite. If you have two lines and both of them are perpendicular to the same plane, plane gamma. So in kind of, yeah, it's kind of a reverse. So you have a plane and you have two perpendiculars, one and another. We have to prove their parallelism. We have to prove that A is parallel to B. All right, so if these are two perpendiculars, these are two bases. What I will do through the point B, I draw a line parallel to A and I assume, let's call it B prime. I assume that this line is not the same as line B. So what happens? From the previous theorem, I know that A and B prime are parallel, which means that from the previous theorem, B prime is perpendicular to gamma. Since A perpendicular to gamma and B prime is parallel to A, then B prime is perpendicular to gamma. Just the previous theorem. So what happens is at this point, point B, I have two perpendiculars B and B prime from the plane gamma. And that is impossible because we have already considered the existence and uniqueness of the perpendicular from the very beginning. Whenever I just introduce one of the first lectures, I think the first lecture about line perpendicular to the plane, I was talking about existence of the perpendicular and their uniqueness. So this contradicts the uniqueness which has been proven in that lecture. So which means that B and B prime are one in the same line, which means that B is parallel to A because B prime is parallel to A by construction. So that's my second theory. Very simple. I just use the first one as the basis. Okay. And the third one, which I wanted to do today is if I have a plane, point M outside of it, and I have two lines A and B. So I have lines A and B. Their intersection is point M, the intersecting M. And I know that A is perpendicular to gamma. So I have a plane gamma. So what I have to prove is, now A and B are two different lines. I have to prove that B is not parallel to A and B is not perpendicular to gamma. So one is a perpendicular and another is something else, which is crossing the gamma. I did not really specify that, but I have to specify that B intersection gamma is not empty. So there is point of intersection because if there is no point of intersection, I can put B parallel. But I'm claiming that B is not parallel to A. I can put parallel to gamma. So all three conditions. B is not parallel to A, B is not parallel to gamma, and B is not perpendicular to gamma. Okay. How can I prove it? Well, some of them are very simple. Since B and A have a common point, B and A cannot be parallel. Since B and gamma have common point, the intersection, B cannot be parallel. Now, how about B being not perpendicular to gamma? How can that be proven? Well, very simply. From the previous theorem, which I have just proven before, if you have two perpendiculars, A is perpendicular by the preposition. And if I assume that B is also perpendicular, then I know that both will be parallel to each other. I just proved it second together. But I know that they are not parallel because they have point of intersection, which means this cannot be perpendicular. You see, I actually proved only one theorem, the first one. And the second and the third were very, very easy consequences from the first. Well, basically, that's it. I do recommend you to try to do exactly the same as I did, just by yourself. And what's very beneficial, I cannot really overstate all the good benefits which you can get from this. If you can write down the proof, whatever I was just explaining, if you can write it down on a piece of paper, rather than just thinking about this, it would greatly improve your understanding of the whole logic. Because whatever you are putting down on paper inculcates into your mind significantly better than if you just think about it. So that's my very, very strong suggestion in this case. And to tell the truth for any other thing, whatever I'm just proving or solving or whatever I'm doing here, if you can try to replicate it after the fact, after you listen to the lecture, just buy yourself on a piece of paper with as much details as possible. For instance, if you remember, I paid attention. Okay, let this be a point of intersection. Is there a point of intersection? Something like this. So try not to miss any piece of the proof. Try to make it in such a way that nobody can answer why. I mean, if you, for instance, said, okay, this is the point of intersection, somebody can ask you, why do you think they intersect? You see? So nothing should be left without the proof and put it in a paper. And, well, let me warn you, it will be a lengthy kind of a composition. But it's worth it, really. That's it. Thanks very much and good luck.